Please help me to prove.

Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \} $then $\displaystyle \Omega$ it's simply connected domain.

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- Nov 19th 2011, 06:03 AMsinichkoComplex Analysis - simply connected domain
Please help me to prove.

Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \} $then $\displaystyle \Omega$ it's simply connected domain.

Thanks! - Nov 19th 2011, 07:42 AMOpalgRe: Complex Analysis - simply connected domain
You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.

Suppose that S is a loop in $\displaystyle \Omega$. The image f(S) of S is a loop in the simply-connected space D. So there is a continuous transformation in D that contracts this loop to a point. The image in of that deformation under the map $\displaystyle f^{-1}$ will be a continuous transformation in $\displaystyle \Omega$ which contracts S to a point S to a point. - Nov 22nd 2011, 11:54 AMsinichkoRe: Complex Analysis - simply connected domain
[QUOTE=Opalg;696240]You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.

Thanks a lot!